p is in the real numbers under the usual norm (and at any rate, all of the p-norms are equivalent in one-dimensional space); also, all of the p-norms have the property that if a sequence converges under one norm, it converges under all of the others, to the same limit.
Also I somewhat naïvely assumed that just as the limit of the p-norm as p->infinity was the sup norm, so the limit of the p-norm as p->0 would be the inf norm (of course for p<1 it's not a real "norm" because it fails the triangle inequality but whatev), but rather it's equal to ∞ outside the coordinate axes and the absolute value of the nonzero coordinate everywhere else; therefore the arclength element is |dx| if dy=0, |dy| if dx=0, and ∞ elsewhere.
Also, the 0-circle turns out to consist of the four points (±1,0) and (0,±1), so its p-arclength is 0 for all p; however, in the sense of "how long it would take to traverse at speed 1 while never leaving it" its length is ∞.
I went ahead and tested this out in Wolfram|Alpha for ever-smaller values of p...
0.5: 14.2832
0.1: 67.7318
0.01: 671.135
0.005: 1341.62
0.002: 3353.07
Below that it was too hard for Wolfram|Alpha to deal with, but basically it shoots to ∞ (and beyond!)
I also tested some high values of p...
3: 6.519535986117990194
4: 6.793869647256916842428
5: 6.9969136160908981518
6: 7.145366367675450102
7: 7.2569521447953687
8: 7.343369831286500739918
9: 7.41207513289697977651
10: 7.467921618445657
11: 7.5141687345623637
After that it got hard for Wolfram|Alpha to perform a proper numerical integration, and for some very high values (like 99 and 100) it returned just 4.; the plots of the relevant integrand for 11 already shows that it's near 4 for nearly the whole domain, so it could be a sampling error, or it could be that the p-circumference does in fact swing down again to approach 4 in the limit...
Anyway, noticing that 2 appears to be the minimum, with minimum value 2π, I tested some values of p near 2...
1.9: 6.28928
1.99: 6.28324
1.999: 6.28319
2: 6.283185307179586476925286766559005768394338798750211641949889184615632812572417997256069650684234136
2.001: 6.28319
2.01: 6.28324
2.1: 6.28818
I have a sneaking suspicion that the minimum value of the p-circumference of the p-circle is indeed found when p=2; in this way our ever-so-homogeneous "natural" Euclidean norm can be said to be optimal.
I also suspect that the p-circumference of the p-circle (under an actual p-norm, which means p is 1 or greater) is the same as the q-circumference of the q-circle when q=1+1/(p-1)...that is, when 1/p+1/q=1 (this interesting relation is commonly seen in statements about L
p spaces).
In particular...
3/2: 6.51954
4/3: 6.793869647256916842427963755382900314043336012387254400780307334760265054775164310119185196453647871061330302026735936783391190405919742048207766394085
5/4: 6.996691
6/5: 7.14537
7/6: 7.25695
8/7: 7.34337
9/8: 7.41208
10/9: 7.46792
11/10: 7.51417
I suspect some integral-substitution would be necessary to
prove this equivalence, but I have a hard time seeing it...but for those who would like to try, substituting p with p/(p-1) in that earlier formula for ds yields (1+(1-x
p/(p-1))
1/(p-1)*x
p/(p-1)²)
1-1/p*dx.
Maybe the substitution u=1-x
p/(p-1) would work, with x=(1-u)
(p-1)/p, so that dx=-(p-1)/p*(1-u)
-1/p*du, and with the interval of integration shifting from (0,1) to (1,0); this transformation (after flipping the interval of integration back to (0,1)) yields ds=(1+(u-u²)
1/(p-1))
1-1/p*(p-1)/p*(1-u)
-1/p*du, which looks interesting in its own right but isn't the same form as the original...
Another interesting idea would be to try to find the p-
area of the p-circle; this seems a bit harder because only the 2-norm is even definable by an inner product:
http://livetoad.org/Courses/Documents/292d/Notes/norms_and_inner_products.pdf
Reading that PDF also led me to wonder whether, if 1/p+1/q=1 and p is at least 1, then the p-
operator norm of a matrix A is equal to the q-operator norm of A* (the
transjugate of A)...
